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UNIVERSAL GRAVITATION Universal Gravitation For any two masses in the universe: F = G m1m2/r2 Celestial G = a constant later evaluated by Cavendish Terrestrial -F +F Sir Isaac Newton m2 m1 1642-1727 r Two people pass in a hall. Find the gravitational force between them. CAVENDISH: MEASURED G Modern value: G = 6.674*10-11 Nm2/kg2 m1 = m2 = 70 kg r=1m m1 m2 r m ⋅m F =G 1 2 2 r F = (6.67 x 10-11 N-m2/kg2)(70 kg)(70 kg)/(1 m)2 F = 3.3 x 10-7 N EarthEarth -Moon Force Practice 1024 kg Mass of Earth: 5.97 x Mass of Moon: 7.35 x 1022 kg Earth-Moon Distance: 3.84 x 108 m What is the force between the earth and the moon? F = (6.67 x 10-11 N m2/ kg2 )(5.97x1024kg)(7.35x1022)/(3.84x108)2 1.98 x 1020 N Page 1 What is the gravitational force of attraction between a 100 kg football player on the earth and the earth? Apparent Weight Definition of Weight Apparent Weight is the normal support force. In an inertial (non-accelerating) frame of reference The weight of an object is the gravitational force the earth exerts on the object. Weight = GMEm/RE2 Weight can also be expressed Weight = mg Combining these expressions mg = GMEm/RE2 • FN = FG What is the weight of a 70 kg astronaut in a satellite with an orbital radius of 1.3 x 107 m? Weight = GMm/r2 Using: G = 6.67 x 10-11 N-m2/kg2 » RE = 6.37*106 m = 6370 km » ME = 5.97 x 1024 kg and M = 5.98 x 1024 kg Weight = 165 N What is the astronaut’s apparent weight? g = GME/RE2 = 9.8 m/s2 The value of the gravitational field strength (g) on any celestial body can be determined by using the above formula. The astronaut is in uniform circular motion about Earth. The net force on the astronaut is the gravitational force. The normal force is 0. The astronaut’s apparent weight is 0. Tides FG by moon on A > FG by moon on B Different distances to moon is dominant cause of earth’s tides Tide Animation FG by moon on B > FG by moon on C Earth-Moon distance: 385,000 km which is about 60 earth radii Sun also produces tides, but it is a smaller effect due to greater Earth-Sun distance. 1.5 x 108 km http://www.youtube.com/watch?v=Ead8d9wVDTQ High high tides; low low tides Low high tides; high low tides Neap Tides Spring Tides Satellite Motion TRMM Tropical Rainfall Measuring Mission The net force on the satellite is the gravitational force. r Fnet= FG Assuming a circular orbit: mac = GmMe/r2 Me m The TRMM orbit is circular and is at an altitude of 218 nautical miles (350 km) and an inclination of 35 degrees to the Equator. The spacecraft takes about 91 minutes to complete one orbit around the Earth. This orbit allows for as much coverage of the tropics and extraction of rainfall data over the 24-hour day (16 orbits) as possible. 2 mM mv =G 2 e r r Note that the satellite mass cancels out. GM e v= r Using M e = 5.97 ×1024 kg For low orbits (few hundred km up) this turns out to be about 8 km/s = 17000 mph Page 2 Geosynchronous Satellite A Colorful Character In order to remain above the same point on the surface of the earth, what must be the period of the satellite’s orbit? What orbital radius is required? Highly accurate data Gave his data to Kepler T = 24 hr = 86,400 s Fnet = FG 2 mM mv =G 2 e r r 4π 2 r 2 GM e = 2 rT 2 r r3 = Using GM eT 2 4π 2 M e = 5.97 × 1024 kg r = 42,000 km = 26,000 mi eccentricity = Orbital Eccentricities Kepler’s First Law Planet The orbit of a planet/comet about the Sun is an ellipse with the Sun's center of mass at one focus PF1 + PF2 = 2a A comet falls into a small elliptical orbit after a “brush” with Jupiter Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune Pluto Notes 0.206 0.007 0.017 0.093 0.048 0.056 0.470 0.009 0.249 Too few observations for Kepler to study Nearly circular orbit Small eccentricity Largest eccentricity among planets Kepler could study Slow moving in the sky Slow moving in the sky Not discovered until 1781 Not discovered until 1846 Not discovered until 1930 eccentricity = c/a Johannes Kepler 1571-1630 or distance between foci divided by length of major axis Kepler’s Second Law Eccentricity Kepler’s Third Law Square of any planet's orbital period (sidereal) is proportional to cube of its mean distance (semi-major axis) from Sun Rav = (Ra + Rp)/2 T2 = K Rav 3 Law of Equal Areas A line joining a planet/comet and the Sun sweeps out equal areas in equal intervals of time Recall from a previous slide the derivation of from Fnet = FG vp va = K = 4π2/GM Planet T (yr) R (AU) T2 R3 Mercury 0.24 0.39 0.06 0.06 Venus 0.62 0.72 0.39 0.37 Ra Rp Page 3 T2 = [4π2/GM]r3 Earth 1.00 1.00 1.00 1.00 Mars Jupiter 1.88 11.9 1.52 5.20 3.53 3.51 142 141 Saturn 29.5 9.54 870 868 He observed it in 1682, predicting that, if it obeyed Kepler’s laws, it would return in 1759. HALLEY’S COMET DISCOVERY OF NEW PLANETS When it did, (after Halley’s death) it was regarded as a triumph of Newton’s laws. Deviations in the orbits of Uranus and Neptune led to the discovery of Pluto in 1930 Newton Simulations & Videos Universal Gravitation Three laws of motion and law of gravitation eccentric orbits of comets cause of tides and their variations the precession of the earth’s axis the perturbation of the motion of the moon by gravity of the sun Solved most known problems of astronomy and terrestrial physics Work of Galileo, Copernicus and Kepler unified. Galileo Galili Nicholaus Copernicus Johannes Kepler 1564-1642 1473-1543 1571-1630 Page 4 http://www.cuug.ab.ca/kmcclary/ http://www.youtube.com/watch?v=fxwjeg_r5Ug http://www.youtube.com/watch?v=AAqSCuHA0j8 http://www.youtube.com/watch?v=0rocNtnD-yI